Question: Mighty Casey hits two baseballs out of the park. The path of the first baseball can be described by the displacement (distance and direction) vector ${\vec{b_1}} = 100 \hat i + 10\hat j$. The path of the second baseball can be described by the displacement vector ${\vec{b_2}} = 90\hat i + (-20)\hat j$. (Values above are given in meters.) How much farther did the first ball travel than the second?
The distance each baseball traveled is the magnitude of its displacement vector. Therefore, the expression $\| {\vec{b_1}} \| - \| {\vec{b_2}} \|$ tells us how much farther the first baseball traveled than the second. $ \| {\vec{b_1}} \| - \| {\vec{b_2}} \| \approx 8.3$ meters Note: We can find the magnitude of any vector $\vec v$ using the Pythagorean theorem $\| \vec v \|^2 = x^2 + y^2$, where $x$ and $y$ are the components of $\vec v$. Let's draw a vector ${\vec{d}}$ to connect the endpoint of the first vector to the endpoint of the second. The magnitude of ${\vec{d}}$ will tell us how far the baseballs are apart. Looking at the diagram, we can see that $ {\vec{b_1}} + {\vec{d}} = {\vec{b_2}}$, so it must also be true that ${\vec{d}} = {\vec{b_2}} - {\vec{b_1}}$. We can find ${\vec{b_2}} - {\vec{b_1}}$ by subtracting the $x$ and $y$ components separately. $\begin{aligned} {\vec{b_2}} - {\vec{b_1}} &= {( 90 \hat i + (-20)\hat j )} - {( 100 \hat i + 10\hat j )} \\\\ &=(90-100) \hat i + (-20-10) \hat j \\\\ &=-10\hat i + (-30)\hat j \end{aligned}$ Applying the Pythagorean theorem to find the magnitude, we get $\| {\vec{d}} \| = \| {\vec{b_2}} - {\vec{b_1}} \| \approx 31.6 \,\text{m}$. The first baseball traveled $8.3$ meters farther than the second baseball. The baseballs are $31.6$ meters apart.